# Simplifying Another Rational Expression

• Sep 10th 2008, 12:38 PM
Cam5
Simplifying Another Rational Expression
Hey there, I can't seem to simplify these rational expressions to save my life.

I know all about bedmas, like terms, factoring, or at least I'd like to think so, but I'm stumped again, here. Is there any oder of steps I can take when simplifying a rational expression to ensure it turns out good?

Here's my attempt:

http://img367.imageshack.us/img367/2...onagainfe6.jpg
http://img367.imageshack.us/img367/r...jpg/1/w216.png

If anyone could solve and explain, that would be super-helpful.

- Cam
• Sep 10th 2008, 12:45 PM
Matt Westwood
You went from:

$\displaystyle \frac{3a (a-2) - (a-4)(a+3)}{(a-2)(a+3)}$

to

$\displaystyle \frac{3a (a-2) - a+4(a+3)}{(a-2)(a+3)}$

This is wrong. You also need to multiply the a of a+4 by the (a+3), you lost some terms.

$\displaystyle \frac{3a (a-2) - (a-4)(a+3)}{(a-2)(a+3)}$

$\displaystyle = \frac{3a (a-2) - (a^2-4a +3a-12)}{(a-2)(a+3)}$ (multiplying out that term before doing anything else)

$\displaystyle = \frac{3a^2 -6 - (a^2-a -12)}{(a-2)(a+3)}$

$\displaystyle = \frac{3a^2 -6 - a^2 + a +12}{(a-2)(a+3)}$ (now you can unpack the minus-thing)

$\displaystyle = \frac{2a^2 + a + 6}{(a-2)(a+3)}$

The thing on top probably factorises. I'll leave it up to you to continue.
• Sep 10th 2008, 01:01 PM
Cam5
Thanks, Matt

$\displaystyle \frac{2a^2 + a + 6}{(a-2)(a+3)}$

The top 'a' variable will factor to be two numbers with a product of 12, and a sum of 1, then.

I can't even think of it...

1 and 12 = 13, so no.
2 and 6 = 8, so no.
3 and 4 = 7, so no.
-3 and 4 = 1, but their product is -12, so that can't be right...
• Sep 10th 2008, 10:27 PM
Matt Westwood
Well if it doesn't factorise, it doesn't. Never mind.